where u is the decision-maker s payoff function over her actions and S is the set of her feasible actions.

Seminars on Mathematics for Economics and Finance Topic 3: Optimization - interior optima 1 Session: 11-12 Aug 2015 (Thu/Fri) 10:00am 1:00pm I. Optimization: introduction Decision-makers (e.g. consumers, firms, governments) in standard economic theory are assumed to be rational. That is, each decision-maker is assumed to have a preference ordering over the outcomes to which her actions lead and to choose an action, among those feasible, that is most preferred according to this ordering. We usually make assumptions that guarantee that a decision-maker s preference ordering is represented by a payoff function (sometimes called utility function), so that we can present the decisionmaker s problem as one of choosing an action, among those feasible, that maximizes the value of this function. That is, we write the decision-maker s problem in the form max u a subject to a S, a where u is the decision-maker s payoff function over her actions and S is the set of her feasible actions. If the decision-maker is a classical consumer, for example, then a is a consumption bundle, u is the consumer s utility function, and S is the set of bundles of goods the consumer can afford. If the decision-maker is a classical firm then a is an input-output vector, u(a) is the 1 This note is largely based on 1) Mathematical methods for economic theory: a tutorial by Martin J. Osborne; and 2) Mathematics Review Course by Paul Schweinzer. 1

profit that action a generates, and S is the set of all feasible input-output vectors (as determined by the firm s technology). Even outside the classical theory, the actions chosen by decision-makers are often modeled as solutions of maximization problems. A firm, for example, may be assumed to maximize its sales, rather than its profit; a consumer may care not only about the bundle of goods she consumes, but also about the bundles of goods the other members of her family consumes, maximizing a function that includes these bundles as well as her own; a government may choose policies to maximize its chance of reelection. In economic theory we sometimes need to solve a minimization problem of the form min u a subject to a S, a We assume, for example, that firms choose input bundles to minimize the cost of producing any given output; an analysis of the problem of minimizing the cost of achieving a certain payoff greatly facilitates the study of a payoff-maximizing consumer. The next three parts of the tutorial develop tools for solving maximization and minimization problems, which are collectively known as optimization problems. This part discusses some basic definitions and a fundamental result regarding the existence of an optimum. The next part, on interior optima, focuses on conditions for solutions that are strictly inside the constraint set S interior solutions. 2

The third part, on equality constraints, discusses the key technique developed by Lagrange for finding the solutions of problems in which S is the set of points that satisfy a set of equations. The last part, on the Kuhn-Tucker conditions for problems with inequality constraints, discusses a set of conditions that may be used to find the solutions of any problem in which S is the set of points that satisfy a set of inequalities. These conditions encompass both the conditions for interior optima and those developed by Lagrange. The last section is a summary of the conditions in the previous parts. II. Optimization: definitions The optimization problems we study take the form maxx f (x) subject to x S where f is a function, x is an n-vector (which we can also write as (x1,..., xn)), and S is a set of n-vectors. We call f the objective function, x the choice variable, and S the constraint set or opportunity set. To be very clear about what we mean by the "max" operator, here is a precise definition. Definition: The value x* of the variable x solves the problem maxx f (x) subject to x S 3

if f (x) f (x*) for all x S. In this case we say that x* is a maximizer of the function f subject to the constraint x S, and that f (x*) is the maximum (or maximum value) of the function f subject to the constraint x S. A minimizer is defined analogously. As an example, both x* and x** are maximizers of f subject to the constraint x S for the function f of one variable in the following figure, and x'' is a minimizer. (The graph of f is horizontal between x3 and x4 and between x5 and x6.) What about the point x'? This point is decisively not a maximizer, because f (x*) > f (x'), for example. But it is a maximum among the points close to it. We call such a point a local maximizer. In the following definition, the distance between two points x and x' is the n square root of x x 2 (sometimes called the Euclidean distance between x and x'). ' i 1 i i Definition: The variable x* is a local maximizer of the function f subject to the constraint x S if there is a number ε > 0 such that f (x) f (x*) for all x S for which the distance between x and x* is at most ε. 4

A local minimizer is defined in the same way. Sometimes we refer to a maximizer as a global maximizer to emphasize that it is not only a local maximizer. Every global maximizer is, in particular, a local maximizer (ε can take any value), and every global minimizer is a local minimizer. For the function f in the figure above we have f (x) f (x') for all x between x1 and x2 (for example), where x1 x' = x2 x', so the point x' is a local maximizer of f (set ε = x1 x'). But note that the point x'' in this figure is also a local maximizer of f, even though it is a global minimizer. The function is constant between x3 and x4. The point x4 is closer to x'' than is the point x3, so we can take the ε in the definition of a local maximizer to be x4 x''. For every point x within the distance ε of x'', we have f (x) = f (x''), so that in particular f (x) f (x''). The point is that the definition of a local maximizer does not require the value of the function at nearby points to be less than f (x*), only less than or equal to f (x*). In economic theory we are almost always interested in global maximizers, not merely local maximizers. For example, the standard theory is that a consumer chooses the bundle she most prefers among all those that are available; she is not satisfied by a bundle that is merely better than the other bundles that are nearby. Similarly, a firm is assumed to choose the input-output vector that maximizes its profit among all those that are feasible; it is not satisfied by an input-output vector that merely yields a higher profit than does similar vectors. 5

Transforming the objective function Let g be a strictly increasing function of a single variable. (That is, if z' > z then g(z') > g(z).) Then the set of solutions to the problem maxx f (x) subject to x S is identical to the set of solutions to the problem maxx g( f (x)) subject to x S. Why? If x* is a solution to the first problem then by definition f (x) f (x*) for all x S. But if f (x) f (x*) then g( f (x)) g( f (x*)), so that g( f (x) g( f (x*)) for all x S. Hence x* is a solution of the second problem. This fact is sometimes useful: for example, if the objective function is f (x1, x2) = x1 α x2 β, it may be easier to work with the logarithm of this function, namely αln x1 + βln x2. Since log z is an increasing function, the set of solutions of the problem in which x1 α x2 β is the objective function is the same as the set of solutions of the problem is which αln x1 + βln x2 is the objective function. 6

Minimization problems Any discussion of maximization problems encompasses minimization problems because of a simple fact: any minimization problem can be turned into a maximization problem by taking the negative of the objective function. That is, the problem minx f (x) subject to x S is equivalent, in particular, has the same set of solutions, as the problem maxx f (x) subject to x S. Thus we can solve any minimization problem by taking the negative of the objective function and apply the results for maximization problems. Existence of an optimum Let f be a function of n variables defined on the set S. The problems we consider take the form maxx f (x) subject to x S where x = (x1,..., xn). 7

Before we start to think about how to find the solution to a problem, we need to think about whether the problem has a solution. Here are some specifications of f and S for which the problem does not have any solution. 1) f (x) = x, S = [0, ) (i.e. S is the set of all nonnegative real numbers). In this case, f increases without bound, and never attains a maximum. 2) f (x) = 1 1/x, S = [1, ). In this case, f converges to the value 1, but never attains this value. 3) f (x) = x, S = (0, 1). In this case, the points 0 and 1 are excluded from S (which is an open interval). As x approaches 1, the value of the function approaches 1, but this value is never attained for values of x in S, because S excludes x = 1. 4) f (x) = x if x < 1/2 and f (x) = x 1 if x 1/2; S = [0, 1]. In this case, as x approaches 1/2 the value of the function approaches 1/2, but this value is never attained, because at x = 1/2 the function jumps down to 1/2. The difficulties in the first two cases are that the set S is unbounded; the difficulty in the third case is that the interval S is open (does not contain its endpoints); and the difficulty in the last case is that the function f is discontinuous. If S is a closed interval [a, b] (where a and b are finite) and f is continuous, then none of the difficulties arise. For functions of many variables, we need to define the concept of a bounded set. 8

Definition: The set S is bounded if there exists a number k such that the distance of every point in S from the origin is at most k. A bounded set does not extend "infinitely" in any direction. Examples: 1) The set [ 1, 100] is bounded, because the distance of any point in the set from 0 is at most 100. The set [0, ) is not bounded, because for any number k, the number 2k is in the set, and the distance of 2k to 0 is 2k which exceeds k. 2) The set {(x, y): x 2 + y 2 4} is bounded, because the distance of any point in the set from (0, 0) is at most 2. 3) The set {(x, y): xy 1} is not bounded, because for any number k the point (2k, 0) is in the set, and the distance of this point from (0, 0) is 2k, which exceeds k. We say that a set that is closed and bounded is compact. Theorem: A set X R n is compact iff it is closed and bounded. The following result generalizes the observations previously. Proposition (Extreme value theorem): A continuous function on a compact set attains both a maximum and a minimum on the set. 9

Note that the requirement of boundedness is on the set, not the function. Requiring that the function be bounded is not enough, as the second example shows: f (x) = 1 1/x, S = [1, ). In this case, f converges to the value 1, but never attains this value. Note also that the result gives only a sufficient condition for a function to have a maximum. If a function is continuous and is defined on a compact set then it definitely has a maximum and a minimum. The result does not rule out the possibility that a function has a maximum and/or minimum if it is not continuous or is not defined on a compact set. (Refer to the section on logic if you are unclear on this point.) Exercise 1: For each of the following functions, determine (i) whether the extreme value theorem implies that the function has a maximum and a minimum and (ii) if the extreme value theorem does not apply, whether the function does in fact have a maximum and/or a minimum. a) x 2 on the interval [ 1,1] b) x 2 on the interval ( 1,1) c) x on the interval [ 1, ) d) f (x) defined by f (x) = 1 if x < 0 and f (x) = x if x 0, on the interval [ 1,1]. e) f (x) defined by f (x) = 1 if x < 0 and f (x) = x if x 0, on the interval (, ). f) f (x) defined by f (x) = x 2 if x < 0 and f (x) = x if x 0 on the interval [ 1,1]. 10

Exercise 2: Does the extreme value theorem imply that the problem maxx pf (x) w x subject to x 0, where x is an n-vector, f is a continuous function, and p > 0 (a scalar) and w > 0 (an n- vector) are parameters, has a solution? If not, specify a function f and values of p and w for which the problem has a solution, if such exist; and specify a function f and values for p and w for which the problem does not have a solution, if such exist. (The problem may be interpreted as the optimization problem of a firm with production function f facing the input price vector w and the price of output p.) 11

III. Necessary conditions for an interior optimum One variable From your previous study of mathematics, you probably know that if the function f of a single variable is differentiable then there is a relationship between the solutions of the problem maxx f (x) subject to x I, where I is an interval of numbers, and the points at which the first derivative of f is zero. What precisely is this relationship? We call a point x such that f '(x) = 0 a stationary point of f. Consider the cases in the three figures. 1) In the left figure, the unique stationary point x* is the global maximizer. 2) In the middle figure, there are three stationary points: x*, x', and x''. The point x* is the global maximizer, while x' is a local (though not global) minimizer and x'' is a local (but not global) maximizer. 3) In the right figure, there are two stationary points: x' and x''. The point x' in neither a local maximizer nor a local minimizer; x'' is a global minimizer. 12

We see that a stationary point is not necessarily a global maximizer, or even a local maximizer, or even a local optimizer of any sort (maximizer or minimizer) (consider x' in the right-hand figure) a global maximizer is not necessarily a stationary point (consider a in the righthand figure). That is, being a stationary point is neither a necessary condition nor a sufficient condition for solving the problem. So what is the relation between stationary points and maximizers? Although a maximizer may not be a stationary point, the only case in which it is not is when it is one of the endpoints of the interval I on which f is defined. That is, any point interior to this interval that is a maximum must be a stationary point. Proposition: Let f be a differentiable function of a single variable defined on the interval I. If a point x in the interior of I is a local or global maximizer or minimizer of f then f '(x) = 0. This result gives a necessary condition for interiors x to be a maximizer (or a minimizer) of f : if it is a maximizer (or a minimizer) and is between a and b then x is a stationary point of f. The condition is obvously not sufficient for a point to be a maximizer the condition is satisfied also, for example, at points that are minimizers. 13

The first-derivative is involved, so we refer to the condition as a first-order condition. Thus among all the points in the interval I, only the endpoints (if any) and the stationary points of f can be maximizers of f. Most functions have a relatively small number of stationary points, so the following procedure to find the maximizers is useful: Procedure for solving a single-variable maximization problem on an interval: Let f be a differentiable function of a single variable and let I be an interval of numbers. If the problem maxx f (x) subject to x I has a solution, it may be found as follows. 1) Find all the stationary points of f (the points x for which f '(x) = 0) that are in the constraint set S, and calculate the values of f at each such point. 2) Find the values of f at the endpoints, if any, of I. The points x you have found at which the value f (x) is largest are the maximizers of f. The variant of this procedure in which the last step involves choosing the points x at which f (x) is smallest may be used to solve the analogous minimization problem. 14

Example One: Consider the problem maxx x 2 subject to x [ 1, 2]. This problem satisfies the conditions of the extreme value theorem, and hence has a solution. Let f (x) = x 2. We have f '(x) = 2x, so the function has a single stationary point, x = 0, which is in the constraint set. The value of the function at this point is f (0) = 0. The values of f at the endpoints of the interval on which it is defined are f ( 1) = 1 and f (2) = 4. Thus the global maximizer of the function on [ 1, 2] is x = 2 and the global minimizer is x = 0. Example Two: Consider the problem maxx x 2 subject to x (, ). This problem does not satisfy the conditions of the extreme value theorem, so that the theorem does not tell us whether it has a solution. Let f (x) = x 2. We have f '(x) = 2x, so that the function has a single stationary point, x = 0, which is in the constraint set. The constraint set has no endpoints, so x = 0 is the only candidate for a solution to the problem. We conclude that if the problem has a solution then the solution is x = 0. In fact, the problem does have a solution: we have f (x) 0 for all x and f (0) = 0, so the solution is indeed x = 0. 15

Example Three: Consider the problem maxx x 2 subject to x (, ). Like the problem in the previous example, this problem does not satisfy the conditions of the extreme value theorem, so that the theorem does not tell us whether it has a solution. Let f (x) = x 2. We have f '(x) = 2x, so that the function has a single stationary point, x = 0, which is in the constraint set. The constraint set has no endpoints, so x = 0 is the only candidate for a solution to the problem. We conclude that if the problem has a solution then the solution is x = 0. In fact, the problem does not have a solution: the function f increases without bound as x increases (or decreases) without bound. Exercise 3: Find all the global maximizers and minimizers of the following functions of a single variable. a. f (x) = 1 x on the interval [0, 1]. b. f (x) = (x 1/3) 2 on the interval [0, 1]. c. f (x) = x 3 2x 2 + x on the interval [0, 2]. 16

Many variables Consider a maximum of a function of two variables. At this maximum the function must decrease in every direction (otherwise the point would not be a maximum!). In particular, the maximum must be a maximum along a line parallel to the x-axis and also a maximum along a line parallel to the y-axis. Hence, given the result for a function of a single variable, at the maximum both the partial derivative with respect to x and the partial derivative with respect to y must be zero. Extending this idea to many dimensions gives us the following result, where f i' is the partial derivative of f with respect to its ith argument. Proposition: Let f be a differentiable function of n variables defined on the set S. If the point x in the interior of S is a local or global maximizer or minimizer of f then f i'(x) = 0 for i = 1,..., n. As for the analogous result for functions of a single variable, this result gives a necessary condition for a maximum (or minimum): if a point is a maximizer then it satisfies the condition. As before, the condition is called a first-order condition. Any point at which all the partial derivatives of f are zero is called a stationary point of f. As for functions of a single variable, the result tells us that the only points that can be global maximizers are either stationary points or boundary points of the set S. Thus 17

the following procedure locates all global maximizers and global minimizers of a differentiable function. Procedure for solving a many-variable maximization problem on a set: Let f be a differentiable function of n variables and let S be a set of n-vectors. If the problem maxx f (x) subject to x S has a solution, it may be found as follows. 1) Find all the stationary points of f (the points x for which f i'(x) = 0 for i = 1,..., n) in the constraint set S and calculate the value of f at each point. 2) Find the largest and smallest values of f on the boundary of S. The points x you have found at which the value of f is largest are the maximizers of f. This method is much less generally useful than the analogous method for functions of a single variable because for many problems finding the largest and smallest values of f on the boundary of S is difficult. For this reason, we devote considerable attention to other, better methods for finding maxima and minima of maximization problems with constraints, in the next two parts. Here are some examples, however, where the method may be fairly easily applied. Example One: Consider the problem maxx,y[ (x 1) 2 (y + 2) 2 ] subject to < x < and < y <. 18

This problem does not satisfy the conditions of the extreme value theorem (because the constraint set is not bounded), so the theorem does not tell us whether the problem has a solution. The first-order conditions are 2(x 1) = 0 2(y + 2) = 0, which have a unique solution, (x, y) = (1, 2). The constraint set has no boundary points, so we conclude that if the problem has a solution, this solution is (x, y) = (1, 2). In fact, the problem does have a solution, because the value of the objective function at (1, 2) is 0, and its value at any point is nonpositive. Example Two: Consider the problems and maxx,y x 2 + y 2 + y 1 subject to x 2 + y 2 1 minx,y x 2 + y 2 + y 1 subject to x 2 + y 2 1. In each case the constraint set, {(x, y): x 2 + y 2 1}, is compact. The objective function is continuous, so by the extreme value theorem, the problem has a solution. We apply the procedure as follows, denoting the objective function by f. 1) We have f 1'(x, y) = 2x and f 2'(x, y) = 2y + 1, so the stationary points are the solutions of 2x = 0 and 2y + 1 = 0. Thus the function has a single stationary point, (x, 19

y) = (0, 1/2), which is in the constraint set. The value of the function at this point is f (0, 1/2) = 5/4. 2) The boundary of the constraint set is the set of points (x, y) such that x 2 + y 2 = 1, as shown in the following figure. 3) Thus for a point (x, y) on the boundary we have f (x, y) = x 2 + 1 x 2 + y 1 = y. We have 1 y 1 on the boundary, so the maximum of the function on the boundary is 1, which is achieved at (x, y) = (0, 1), and the minimum is 1, achieved at (x, y) = (0, 1). 4) Looking at all the values we have found, we see that the global maximum of f is 1, achieved at (0, 1), and the global minimum is 5/4, achieved at (0, 1/2). Notice that the reasoning about the behavior of the function on the boundary of the constraint set is straightforward in this example because we are able easily to express the value of the function on the boundary in terms of a single variable (y). In many other problems, doing so is not easy. 20

Exercise 4: A consumer has the utility function x1x2, has income y > 0, and faces the prices p1 > 0 and p2 > 0. She is required to spend all her income. Formulate her problem as a constrained maximization problem in which the variables are x1 and x2 and the constraints are the budget constraint, x1 0, and x2 0. Transform this problem into a maximization problem in the single variable x1, which is constrained to be in an interval, by isolating x2 in the budget constraint and substituting it into the utility function. Solve this problem using the procedure in the previous lecture note, thus finding the bundle (x1, x2) that maximizes the consumer's utility subject to her budget constraint. 21

IV. Local optima One variable Occasionally we are interested only in the local maximizers or minimizers of a function. We may be able to tell whether a stationary point is a local maximum, a local minimum, or neither by examining the second derivative of the function at the stationary point. Proposition (Second-order conditions for optimum of function of one variable): Let f be a function of a single variable with continuous first and second derivatives, defined on the interval I. Suppose that x* is a stationary point of f in the interior of I (so that f '(x*) = 0). 1) If f "(x*) < 0 then x* is a local maximizer. 2) If x* is a local maximizer then f "(x*) 0. 3) If f "(x*) > 0 then x* is a local minimizer. 4) If x* is a local minimizer then f "(x*) 0. If f "(x*) = 0 then we don't know, without further investigation, whether x* is a local maximizer or local minimizer of f, or neither (check the functions x 4, x 4, and x 3 at x = 0). In this case, information about the signs of the higher order derivatives may tell us whether a point is a local maximum or a local minimum. In practice, however, these conditions are rarely useful. 22

Many variables As for a function of a single variable, a stationary point of a function of many variables may be a local maximizer, a local minimizer, or neither, and we may be able to distinguish the cases by examining the second-order derivatives of the function at the stationary point. Let (x0, y0) be a stationary point of the function f of two variables. Suppose it is a local maximizer. Then certainly it must be a maximizer along the two lines through (x0, y0) parallel to the axes. Using the theory for functions of a single variable, we conclude that f 11"(x0, y0) 0 and f 22"(x0, y0) 0, where f ij" denotes the second partial derivative of f with respect to its ith argument, then with respect to its jth argument. However, even the variant of this condition in which both inequalities are strict is not sufficient for (x0, y0) to be a maximizer, as the following example shows. Example: Consider the function f (x, y) = 3xy x 2 y 2. The first-order conditions are f 1'(x, y) = 3y 2x = 0 f 2'(x, y) = 3x 2y = 0 so that f has a single stationary point, (x, y) = (0, 0). Now, f 11"(0, 0) = 2 0 23

f 22"(0, 0)= 2 0. But (0, 0) is not a local maximizer: at (0, 0) the value of the function is 0, but at (ε, ε) with ε > 0 the value of the function is 3ε 2 ε 2 ε 2 = ε 2, which is positive (and hence exceeds f (0, 0) = 0) no matter how small ε is. This example shows that we cannot determine the nature of a stationary point of a function f of two variables by looking only at the partial derivatives f 11" and f 22" at the stationary point. The next result gives a condition that involves the definiteness of the Hessian of the function, and thus all the cross-partials. The result assumes that all the second-order partial derivatives f ij" are continuous for all x in some set S, so that by Young's theorem we have f ij"(x) = f ji"(x) for all x S, and hence the Hessian is symmetric. Proposition (Second-order conditions for optimum of function of many variables): Let f be a function of n variables with continuous partial derivatives of first and second order, defined on the set S. Suppose that x* is a stationary point of f in the interior of S (so that f i'(x*) = 0 for all i). 1) If H(x*) is negative definite then x* is a local maximizer. 2) If x* is a local maximizer then H(x*) is negative semidefinite. 3) If H(x*) is positive definite then x* is a local minimizer. 24

4) If x* is a local minimizer then H(x*) is positive semidefinite. An implication of this result is that if x* is a stationary point of f then 1) if H(x*) is negative definite then x* is a local maximizer 2) if H(x*) is negative semidefinite, but neither negative definite nor positive semidefinite, then x* is not a local minimizer, but might be a local maximizer 3) if H(x*) is positive definite then x* is a local minimizer 4) if H(x*) is positive semidefinite, but neither positive definite nor negative semidefinite, then x* is not a local maximizer, but might be a local minimizer 5) if H(x*) is neither positive semidefinite nor negative semidefinite then x* is neither a local maximizer nor a local minimizer. For a function f of two variables, the Hessian is * * f 11( x ) f 12( x ) * * f 21( x ) f 22( x ) This matrix is negative definite if f 11"(x*) < 0 and H(x*) > 0. (These two inequalities imply that f 22"(x*) < 0.) Thus the extra condition, in addition to the two conditions f 11"(x*) < 0 and f 22"(x*) < 0 considered originally, for x* to be a local maximizer is f 11"(x*) f 22"(x*) f 21"(x*) f 12"(x*) > 0. Similarly, a sufficient condition for a stationary point x* of a function of two variables to be a local maximizer are f 11"(x*) > 0 and H(x*) > 0 (which imply that f 22"(x*) > 0). 25

In particular, if, for a function of two variables, H(x*) < 0, then x* is neither a local maximizer nor a local minimizer. (Note that this condition is only sufficient, not necessary.) A stationary point that is neither a local maximizer nor a local minimizer is called a saddle point. Examples are the point (0, 0) for the function f (x, y) = x 2 y 2 and the point (0, 0) for the function f (x, y) = x 4 y 4. In both cases, (0, 0) is a maximizer in the y direction given x = 0 and a minimizer in the x direction given y = 0; the graph of each function resembles a saddle for a horse. Note that not all saddle points look like saddles. For example, every point (0, y) is a saddle point of the function f (x, y) = x 3. From the results above, a sufficient, though not necessary, condition for a stationary point x* of a function f of two variables to be a saddle point is H(x*) < 0. Example 1: Consider the function f (x, y) = x 3 + y 3 3xy. The first-order conditions for an optimum are 3x 2 3 y = 0 3y 2 3x = 0. Thus the stationary points satisfy y = x 2 = y 4, so that either (x, y) = (0, 0) or y 3 = 1. So there are two stationary points: (0, 0), and (1, 1). Now, the Hessian of f at any point (x, y) is 6x 3 H ( x, y) 3 6y 26

Thus H(0, 0) = 9, so that (0, 0) is neither a local maximizer nor a local minimizer (i.e. is a saddle point). We have f 11"(1, 1) = 6 > 0 and H(1, 1) = 36 9 > 0, so that (1, 1) is a local minimizer. Example 2: Consider the function f (x, y) = 8x 3 + 2xy 3x 2 + y 2 + 1. We have f 1'(x, y) = 24x 2 + 2y 6x f 2'(x, y) = 2x + 2y. So the Hessian is 48x 6 2 2 2 To find the stationary points of the function, solve the first-order conditions. From the second equation have y = x; substituting this into first equation we find that 24x 2 8x = 8x(3x 1) = 0. This equation has two solutions, x = 0 and x = 1/3. Thus there are two solutions of the first-order conditions: (x*, y*) = (0, 0) and (x**, y**) = (1/3, 1/3). Now look at the second-order condition. We have f 11"(x, y) = 48x 6, f 22"(x, y) = 2, and f ''12(x, y) = 2. Now look at each stationary point in turn: 1) (x*, y*) = (0, 0) We have f 11"(0, 0) = 6 < 0 and 27

f 11"(0, 0) f 22"(0, 0) ( f 12"(0, 0)) 2 = 16 < 0. So (x*, y*) = (0, 0) is neither a local maximizer nor a local minimizer (i.e. it is a saddle point). 2) (x**, y**) = (1/3, 1/3) We have f 11"(1/3, 1/3) = 10 > 0 and f 11"(1/3, 1/3) f 22"(1/3, 1/3) ( f 12"(1/3, 1/3)) 2 = 96/3 16 = 16 > 0. So (x**, y**) = (1/3, 1/3) is a local minimizer. The minimum value of the function is f (1/3, 1/3) = 23/27. Exercise 5: Consider the function f (x, y) = (x 2) 4 + (y 3) 4. a) Show that this function has a minimum at (x, y) = (2, 3) (without using any calculus). b) Find all the solutions of the first-order conditions. c) Is the Hessian of f positive definite at any solution of the first-order conditions? 28

V. Conditions under which a stationary point is a global optimum One variable Let f be a differentiable concave function. Then for every point x, no point on the graph of f lies above the tangent to f at x. Thus if f '(x) = 0 (i.e. if x is a stationary point of f ), the point x is a (global) maximizer of f. Similarly, a differentiable convex function lies nowhere below any of its tangents. Thus we have the following result. Proposition: Let f be a differentiable function defined on the interval I, and let x be in the interior of I. Then 1) if f is concave then x is a global maximizer of f in I if and only if x is a stationary point of f 2) if f is convex then x is a global minimizer of f in I if and only if x is a stationary point of f. Now, a twice-differentiable function is concave if and only if its second derivative is nonpositive (and similarly for a convex function), so we deduce that if f is a twicedifferentiable function defined on the interval I and x is in the interior of I then 1) f "(z) 0 for all z I [x is a global maximizer of f in I if and only if f '(x) = 0] 2) f "(z) 0 for all z I [x is a global minimizer of f in I if and only if f '(x) = 0]. 29

Example 1: Consider the problem maxx -x 2 subject to x [ 1, 1]. The function f is concave; its unique stationary point is 0. Thus its global maximizer is 0. Example 2: A competitive firm pays w for each unit of an input. It obtains the revenue p for each unit of output that it sells. Its output from x units of the input is x. For what value of x is its profit maximized? The firm s profit is p x wx. The derivative of this function is (1/2)px 1/2 w, and the second derivative is (1/4)px 3/2 0, so the function is concave. So the global maximum of the function occurs at the stationary point. Hence the maximizer solves (1/2)px 1/2 w = 0, so that x = (p/2w) 2. What happens if the production function is x 2? Many variables A differentiable concave function of many variables always lies below, or on, its tangent plane, and a differentiable convex function always lies above, or on, its tangent plane. Thus as for functions of a single variable, every stationary point of a concave function of many variables is a maximizer, and every stationary point of a convex function of many variables is a minimizer, as the following result claims. 30

Proposition: Suppose that the function f has continuous partial derivatives in a convex set S and let x be in the interior of S. Then 1) if f is concave then x is a global maximizer of f in S if and only if it is a stationary point of f 2) if f is convex then x is a global minimizer of f in S if and only if it is a stationary point of f. As in the case of functions of a single variable, we can combine this result with a previous result characterizing twice-differentiable concave and convex functions to conclude that if f is a function with continuous partial derivatives of first and second order on a convex set S, and x is in the interior of S, then 1) H(z) is negative semidefinite for all z S [x is a global maximizer of f in S if and only if x is a stationary point of f ] 2) H(z) is positive semidefinite for all z S [x is a global minimizer of f in S if and only if x is a stationary point of f ], where H(x) denotes the Hessian of f at x. Be sure to notice the difference between the form of this result and that of the the result on local optima. To state briefly the results for maximizers together: Sufficient conditions for local maximizer: if x* is a stationary point of f and the Hessian of f is negative definite at x* then x* is a local maximizer of f 31

Sufficient conditions for global maximizer: if x* is a stationary point of f and the Hessian of f is negative semidefinite for all values of x then x* is a global maximizer of f. Example 1: Consider the function f (x, y) = x 2 + xy + 2y 2 + 3, defined on the domain (, ). We have f 'x(x, y) = 2x + y f 'y(x, y) = x + 4y so that the function has a single stationary point, (x, y) = (0, 0). We have f ''xx(x, y) = 2, f ''yy(x, y) = 4, and f ''xy(x, y) = 1, so the Hessian of f is 2 1 1 4 which is positive definite for all values of (x, y). (The matrix is independent of (x, y).) Hence f is convex (in fact, strictly convex). Thus the global minimizer of the function is (0, 0); the minimum value of the function is 3. 32

Example 2: Consider the function f (x, y) = x 4 + 2y 2, defined on the domain (, ). We have f 'x(x, y) = 4x 3 f 'y(x, y) = 4y, so that the function has a single stationary point, (x, y) = (0, 0). We have f ''xx(x, y) = 12x 2, f ''yy(x, y) = 4, and f ''xy(x, y) = 0, so the Hessian of f is 2 12x 0 0 4 This matrix is positive semidefinite for all values of (x, y), so that (0, 0) is the unique global minimizer of the function. 33

Example 3: Consider the function f (x, y) = x 3 + 2y 2, defined on the domain (, ). We have f 'x(x, y) = 3x 2 f 'y(x, y) = 4y, so that the function has a single stationary point, (x, y) = (0, 0). We have f ''xx(x, y) = 6x, f ''yy(x, y) = 4, and f ''xy(x, y) = 0, so the Hessian of f is 6x 0 0 4 At (x, y) = (0, 0) this matrix is positive semidefinite, but not positive definite. Thus we cannot tell from this analysis whether (0, 0) is a local maximizer or local minimizer, or neither. (In fact, it is neither: for all ε > 0 the function is positive at (ε, 0) and negative at ( ε, 0).) At other values of (x, y) the matrix is not positive semidefinite (for example, if x < 0 the matrix is not positive semidefinite), so the function is not concave. Thus we cannot conclude that (0, 0) is either a global maximizer or minimizer. In summary, we can conclude from this analysis only that if the function has a minimizer, then this minimizer is (0, 0), and if it has a maximizer, then this maximizer is (0, 0). In fact, the function does not have either a minimizer or maximize. It is an arbitrarily large negative number when x is an arbitrarily large negative number, and is an arbitrarily large positive number when y is an arbitrarily large positive number. 34

Example 4: Consider a firm with the production function f, defined over vectors (x1,..., xn) of inputs. Assume that f is concave. The firm s profit function is ( x,, x ) pf ( x,, x ) w x 1 n 1 n j 1 j j where p is the price of the output of the firm and wj is the price of the jth input. The second term in the profit function is linear, hence concave, and so the function π is concave. So the input bundle (x1*,..., xn*) > 0 maximizes the firm's profit if and only if p f 'j(x*) = wj for all j (i.e. the value of the marginal product of each input j is equal to its price). n Exercise 6: 1. A politician chooses the number of hours h 0 of advertising that she buys; the cost of h hours is c(h), where c is a convex function. She wishes to maximize g(π(h)) c(h), where π(h) is her probability of winning when she buys h hours of advertising. Assume that π is a concave function, and g is an increasing and concave function. a. Write down the first-order condition for an interior solution (i.e. a solution h* with h* > 0) of this problem. b. What is the relation (if any) between an interior solution of the first-order condition and the solution of the problem? 35

2. A firm sells goods in two markets. In each market i the price is fixed at pi, with pi > 0. The amount of the good the firm sells in each market depends on its advertising expenditure in both markets. If it spends ai on advertising in market i, for i = 1, 2, its sales in market 1 are f (a1, a2) and its sales in market 2 are g(a1, a2), where f and g are twice differentiable functions. The firm s cost of production is zero. Thus its profit is: p1 f (a1,a2) + p2g(a1, a2) a1 a2. It chooses (a1, a2) to maximize this profit. a. What are the first-order conditions for the firm s optimization problem? b. Suppose that (a1*, a2*) maximizes the firm s profit and ai* > 0 for i = 1, 2. Does (a1*, a2*) necessarily satisfy the first-order conditions? c. Suppose that (a1*, a2*) satisfies the first-order conditions and ai* > 0 for i = 1, 2. Under what condition is (a1*, a2*) a local maximizer of the firm s profit? d. Suppose that (a1*, a2*) satisfies the first-order conditions and ai* > 0 for i = 1, 2. Under what conditions on the functions f and g is (a1*, a2*) necessarily a global maximizer of the firm s profit? 36